NinetynineCAT 2025 Slot 2 QA Q6 Geometry
Question 6 of the Quantitative Ability (QA) section, CAT 2025 Slot 2, with CAT's official answer and a worked solution.
Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is
- A5 : 24
- B7 : 24
- C6 : 25
- D6 : 19
Answer: (A) 5 : 24 Geometry MCQ
Solution
- Take side 1 and centre at the origin: \(A(1,0)\), \(B(\frac12,\frac{\sqrt3}{2})\), \(C(-\frac12,\frac{\sqrt3}{2})\), \(D(-1,0)\).
- Midpoints: \(P(\frac34,\frac{\sqrt3}{4})\), \(Q(-\frac34,\frac{\sqrt3}{4})\). So \(PQ = \frac32\), \(BC = 1\), both horizontal, height \(\frac{\sqrt3}{4}\).
- Trapezium area \(= \frac12(\frac32 + 1)\cdot\frac{\sqrt3}{4} = \frac{5\sqrt3}{16}\).
- Hexagon area \(= \frac{3\sqrt3}{2} = \frac{24\sqrt3}{16}\). Ratio \(= 5 : 24\).
Shortcut. PQ is the midline between BC (length s) and the long diagonal AD (2s), so it is 1.5s, at half the height.